4. ) For each of the statements below (which all happen to be false), look at the "proofs" that have been provided and explain in one or two sentences what error has been mnade by the author. (Note: We are not looking for picky details about writing style here, but actual major problems with the proofs.) (a) Statement: For all odd integers m and n, mtn is an integer. "Proof": Let m = 3 and let n = 5. By definition of odd, 3 is odd because 3 = 2(1) + 1, and 1 is an integer. Similarly, 5 is odd because 5 = 2(2) + 1, and 2 is an integer. We wish to prove that is an integer. 3+5 * (by substitution) 32 (by algebra) (by algebra) = 7 (by algebra) Since 7 is an integer, we have shown what we were required to show. O (b) Statement: For all integers m, if 6m – 10 is even, then m is odd. "Proof": Let m be any odd integer. WWe wish to prove that 6m- 10 is even. In other words, we wish to show that 6m – 10 = 2k for some integer k. Since m is odd, we know that m = 2p + 1 for some integer p (by definition of odd). Now, 6m - 10 = 6(2p + 1) – 10 = 12p + 6 – 10 = 12p – 4 = 2(6p – 2) (by substitution) (by algebra) (by algebra) (by algebra) We know that 6p - 2 is an integer, since it is the product and difference of integers (6, p, 2). 10 = 2k for an integer k. By definition of even, 6m - 10 Therefore, we have shown that 6m is even. O
4. ) For each of the statements below (which all happen to be false), look at the "proofs" that have been provided and explain in one or two sentences what error has been mnade by the author. (Note: We are not looking for picky details about writing style here, but actual major problems with the proofs.) (a) Statement: For all odd integers m and n, mtn is an integer. "Proof": Let m = 3 and let n = 5. By definition of odd, 3 is odd because 3 = 2(1) + 1, and 1 is an integer. Similarly, 5 is odd because 5 = 2(2) + 1, and 2 is an integer. We wish to prove that is an integer. 3+5 * (by substitution) 32 (by algebra) (by algebra) = 7 (by algebra) Since 7 is an integer, we have shown what we were required to show. O (b) Statement: For all integers m, if 6m – 10 is even, then m is odd. "Proof": Let m be any odd integer. WWe wish to prove that 6m- 10 is even. In other words, we wish to show that 6m – 10 = 2k for some integer k. Since m is odd, we know that m = 2p + 1 for some integer p (by definition of odd). Now, 6m - 10 = 6(2p + 1) – 10 = 12p + 6 – 10 = 12p – 4 = 2(6p – 2) (by substitution) (by algebra) (by algebra) (by algebra) We know that 6p - 2 is an integer, since it is the product and difference of integers (6, p, 2). 10 = 2k for an integer k. By definition of even, 6m - 10 Therefore, we have shown that 6m is even. O
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:4.
-) For each of the statements below (which all happen to be false), look at the "proofs" that
have been provided and explain in one or two sentences what error has been made by the author.
(Note: We are not looking for picky details about writing style here, but actual major problems with
the proofs.)
(a) Statement: For all odd integers m and n,
mtn is an integer.
"Proof": Let m = 3 and let n = 5. By definition of odd, 3 is odd because 3 = 2(1) + 1, and 1
is an integer. Similarly, 5 is odd because 5 = 2(2) + 1, and 2 is an integer.
We wish to prove that ": is an integer.
min = 315 (by substitution)
2 (by algebra)
(by algebra)
= 7
(by algebra)
Since 7 is an integer, we have shown what we were required to show. O
(b) Statement: For all integers m, if 6m – 10 is even, then m is odd.
"Proof": Let m be any odd integer. We wish to prove that 6m - 10 is even. In other words,
we wish to show that 6m – 10 = 2k for some integer k.
Since m is odd, we know that m = 2p + 1 for some integer p (by definition of odd).
Now, 6m- 10 = 6(2p + 1) – 10
= 12p + 6 – 10
12р — 4
3D 2(бр — 2)
(by substitution)
(by algebra)
(by algebra)
(by algebra)
We know that 6p – 2 is an integer, since it is the product and difference of integers (6, p, 2).
Therefore, we have shown that 6m - 10 = 2k for an integer k. By definition of even, 6m – 10
is even. O
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